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### The Square Root of n!

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Date: 10/14/98 at 08:14:43
From: preston Johnson
Subject: Algebra, square roots, and factorials

For what natural numbers n is the square root of n! an integer?

I tried putting different numbers in for n. If I did the problem at all
problem or how to set up an equation. If you could help, that would be
great.
```

```
Date: 10/14/98 at 09:30:41
From: Doctor Mitteldorf
Subject: Re: Algebra, square roots, and factorials

Dear Preston,

Among factorials, only 0! and 1! are perfect squares. Your result for
23 and 24 is probably due to the round-off error in your calculator or
computer.

To prove that a factorial bigger than 1 can't be a perfect square,
first think about breaking down the factorial into prime factors. For
example, 4! = 24, and 24 = 2*2*2*3. In order for any number to be a
perfect square, it must contain an even number of each prime factor.
Now think about the largest prime less than or equal to n. Call that
number p. That number appears only once in the list of factors
1*2*3*4*5*...*p*...*n, so it can't appear an odd number of times
unless 2*p is also less than n.

Now there's a semi-famous theorem called Bertrand's Postulate, which
says that there is always a prime number between n and 2n. This is just
what you need to show that 2p isn't less than n - because if it were,
there would have to be a larger prime than p that is also less than n.
The proof of Bertrand's postulate isn't easy, but you can read it at:

http://mathforum.org/dr.math/problems/bolton.8.1.96.html

You may also be interested in programming in a language that allows
you to work with numbers that have more than 20 or so digits, which
most calculators and computer programing languages can't do.  Look at
the program Mathematica - it contains wonders!

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

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